Geometrical optics - lens floating in mercury

[Froskoy]

Homework Statement

Consider a thin bi-convex lens with refractive index $n$ which has spherical surfaces with equal radii of curvature r and a measured focal length $f$. The lens floats horizontally on the surface of liquid mercury so that its lower surface effectively becomes a spherical mirror. A point object on its optical axis is then found to coincide with its image when it is at a height $u$ above the lens. Determine $r$ and $n$.

Homework Equations

$$\frac{n_2}{v}+\frac{n_1}{u}=\frac{n_2-2_1}{R}$$

$$\frac{1}{u}+\frac{1}{v}=\frac{1}{f}$$

The Attempt at a Solution

Here is my ray diagram for the situation:

http://dl.dropbox.com/u/50942032/ray-diagram.JPG

The distance $v$ represents where the image would be formed if there was no lower surface to the lens (just refraction at a curved surface) so that the thin lens equation can be used. I've marked in some angles on the diagram:

$\alpha$ represents the angle between the incident ray from the point object and the vertical.

The other angles are angles which I've tried to use in formulating a solution - no success so far.

I'm thinking that one approach would be to find the coordinates of intersection with the lower surface of the lens and then use geometry to find the coordinates of intersection with the top and then set up some simultaneous equations and solve them, but I haven't had any success with this approach.

I'm guessing the small angle approximation can be used to approximate $\sin\theta\approx\theta$ in Snell's law and [itex]\tan\theta\approx\theta[\itex] in any geometry.

The light will be refracted when it reaches the top surface of the lens (can apply Snell's law here), and then reflected on the bottom surface such that the angle of incidence equals the angle of reflection. Would it be possible to somehow use the refractive index of mercury (=1.000933) to work out the conditions for the critical angle - would this even help?

I'm really stuck with this and would very much appreciate any help.

With many thanks,

Froskoy.

 

[mark726]

Dear Froskoy,

Thank you for your detailed post and for providing a diagram to help explain the situation. I understand that you are trying to determine the radius of curvature and refractive index of the lens based on its focal length and the height at which the point object and image coincide.

One approach to solving this problem is to use the thin lens equation, which relates the object distance (u), image distance (v), and focal length (f) of a thin lens. This equation assumes that the lens is thin and that light rays passing through it are only slightly deviated. However, in this case, we have a bi-convex lens with a lower surface that acts as a spherical mirror, so the thin lens equation alone may not be sufficient.

Instead, we can use a combination of the thin lens equation and the mirror equation, which relates the object distance (u), image distance (v), and focal length (f) of a spherical mirror. Since the lower surface of the lens effectively acts as a spherical mirror, we can use this equation to find the image distance (v) when the object distance (u) is known.

Using the thin lens equation, we can write:

1/f = 1/u + 1/v

Using the mirror equation, we can write:

1/f = 1/u + 1/v'

where v' is the image distance when the lower surface acts as a spherical mirror.

Since the point object and image coincide when the lens is at a height u above the lens, we can equate v and v':

v = v'

Combining these equations, we get:

1/u + 1/v = 1/u + 1/v'

Simplifying, we get:

1/v = 1/v'

This means that the image distance when the lower surface acts as a spherical mirror is the same as the image distance when the lens is considered to be a thin lens. This also implies that the focal length of the lens is equal to the focal length of the spherical mirror, which is equal to half the radius of curvature.

Therefore, we can write:

f = R/2

Substituting this into the thin lens equation, we get:

1/R = 1/u + 1/v

Solving for R, we get:

R = u*v/(u + v)

Now, we can use the given focal length (f) and the height at which the point object and image coincide